UVALive 2052 Number Steps【简单模拟】水题

本文链接：https://blog.youkuaiyun.com/mengxiang000000/article/details/51985264

本程序通过预先填充一个二维数组的方式，模拟在一个平面上按照特定规律排列数字的过程，并根据输入的坐标(x, y)，返回该位置上的数字。若该位置未分配数字，则输出提示信息。

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A - Number Steps

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Submit Status

Description

Starting from point (0,0) on a plane, we have written all non-negative integers 0,1,2,... as shown in the figure. For example, 1, 2, and 3 has been written at points (1,1), (2,0), and (3,1) respectively and this pattern has continued.

You are to write a program that reads the coordinates of a point (x, y), and writes the number (if any) that has been written at that point. (x, y) coordinates in the input are in the range 0...5000.

Input

The first line of the input is N, the number of test cases for this problem. In each of the N following lines, there is x, and y representing the coordinates (x, y) of a point.

Output

For each point in the input, write the number written at that point or write "No Number" if there is none.

Sample Input

4 2

6 6

3 4

Sample Output

No Number

题目大意：按照图中所示方式排布数字，问点x，y是否有数字，如果有，输出，否则输出没有。

思路：

简单模拟。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int a[5005][5005];
void init()
{
    int x=1,y=1;
    int tmp=1;
    int d=3;
    memset(a,0,sizeof(a));
    while(1)
    {
        a[x][y]=tmp;
        x++;y++;
        tmp+=d;
        if(d==3)d=1;
        else d=3;
        if(x>5002||y>5002)break;
    }
    x=2;y=0;
    tmp=2;
    d=1;
    while(1)
    {
        a[x][y]=tmp;
        x++;y++;
        tmp+=d;
        if(d==3)d=1;
        else d=3;
        if(x>5002||y>5002)break;
    }
}
int main()
{
    init();
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        if(x==0&&y==0)
        {
            printf("0\n");
            continue;
        }
        if(a[x][y]==0)
        {
            printf("No Number\n");
        }
        else printf("%d\n",a[x][y]);
    }
}